Question

Factorise:
(i) $x^3-2x^2-x+2$
(ii) $x^3-3x^2-9x-5$
(iii) $x^3+13x^2+32x+20$
(iv) $2y^3+y^2-2y-1$

Answer 1

(i) Given $x^3-2x^2-x+2$
$=x^2(x-2)-1(x-2)$ [ taking $x^2$ and $-1$ as common]
$=(x-2)(x^2-1)$ [ Taking $x-2$ as a common]
$=(x-2)(x+1)(x-1)$ [ since, $a^2-b^2=(a-b)(a+b)$]

(ii) Given $x^3-3x^2-9x-5$
Here $-3x^2=-5x^2+2x^2$

$=x^3-5x^2+2x^2-10x+x-5$

$=x^2(x-5)+2x(x-5)+1(x-5)$ [taking $x^2,2x$ and $1$ as common]

$=(x-5)(x^2+2x+1)$ [taking $(x-5)$ as common]
Here, $x^2+2x+1$ is in the form of $a^2+2ab+b^2=(a+b{)}^2$
So, $a=x$ and $b=1$
Thus, $x^2+2x+1=(x+1{)}^2$

Hence, $x^3-3x^2-9x-5=(x-5)(x+1)(x+1)$

(iii) Given $x^3+13x^2+32x+20$

Here, $13x^2=2x^2+11x^2,32x=22x+10x$
Thus, $x^3+13x^2+32x+20=x^3+2x^2+11x^2+22x+10x+20$
$=x^2(x+2)+11x(x+2)+10(x+2)$ [taking $x^2,11x$ and $10$ as common]

$=(x+2)(x^2+11x+10)$

$=(x+2)(x^2+x+10x+10)$ [ Here $11x=10x+1x$]

$=(x+2)\left[x\right(x+1)+10(x+1\left)\right]$ [ Taking $x,10$ as a common]

$=(x+2)(x+1)(x+10)$

(iv) Given $2y^3+y^2-2y-1$
Here, $y^2=2y^2-y^2$ and $-2y=-y-y$

Thus, $2y^3+y^2-2y-1=2y^3+2y^2-y^2-y-y-1$

$=2y^2(y+1)-y(y+1)-1(y+1)$ [ Taking $2y^2,-y$ and $-1$ as common]

$=(y+1)\{2y^2-y-1\}$

$=(y+1)\{2y^2-2y+y-1\}$ [ Here $-y=-2y+y$]

$=(y+1)\left\{2y\right(y-1)+1(y-1\left)\right\}$ [ taking $2y$ and $1$ as a common]

$=(y-1)(y+1)(2y+1)$